Is strength of pi bonds greater than sigma bond of nitrogen molecule?

Question

I am referring to molecular orbital (MO) theory. In the nitrogen molecule (and elements of lower atomic number) $\textrm{sp}$ mixing occurs and the sigma orbital set is raised above the pi orbital set. Shouldn't that mean that pi bond is stronger than sigma bond in this case?

Answer 1

You can actually calculate orbital "energies" with MO theory. The resulting "energies" for $\ce{N2}$ are the following (calculated with molpro and CASSCF(10,8)/aug-cc-pVTZ):

Orbital "energies" in atomic units

orbital	energy/a.u.
1σg	-1.11
2σg	-0.99
1σu	-0.77
πu	-0.59
πg	0.29
2σu	1.22

As you can see, the "energy" of the π orbitals is above the "energy" of the 2σ_g, which is even below the 1σ_u "energy". This disagrees with virtually every MO scheme of $\ce{N2}$ you can find. So why is that?

In the end, MO schemes are used in what is called "qualitative molecular orbital theory" (QMO theory). There are usually no underlying calculations (or the calculations were performed in the 60s) and the schemes are adapted in order to explain and understand the stability and reactivity of molecules.

On the other hand, there is computational MO theory, which could in principle be used to improve these diagrams. Why is this not done? In the end, electronic states of a molecule exist and have energies, while orbitals don't. They are not in any way 'real' or observable and are in general best understood as a tool for building approximate wave functions. The orbital "energies" are eigenvalues of the so-called Fock operator, but they should not be given too much attention. For the most accurate wave functions (the so-called full-CI wave functions), the orbital energies are actually arbitrary.

In the end, MO schemes are a tool used to understand and simplify the discussion of chemical bonding. If they cease to be simple, they should not be used.

As for the discussion of the π bond energy vs. the σ bond energy: again, there are not really separate π and σ bonds. However, if you want to calculate some energies aligning with these concepts, valence bond theory is better at that than MO theory.

Answer 2

There is a similar question on StackExchange from 6 years ago (Ref 1), from which I have taken Diagram #1:

DIAGRAM #1

And a quote from the answer: "Many of the diagrams (on the internet) are different one from another - that's a tip off that s-p mixing is not something that can always be predicted in advance." Something like Diagram #2 below, which implies a very different degree of hybridization (none). Just to be clear, "N₃" means the nitrogen molecule (which is N₂).

DIAGRAM #2

The four orbitals (2 sigma atomic and 2 pi atomic) interact and separate into 4 sigma bonds, two bonding and two antibonding. The lowest three sigma MOs are filled in N₂. The take-away from both diagrams is that the three lowest energy sigma MOs are energetically more stable than the atomic s and p orbitals, even tho formally, the uppermost filled sigma level is antibonding.

The degree of splitting is dependent on the amount of hybridization entered into the calculation, and if that is small, the actual energy levels will resemble Diagram #2, and if the amount of hybridization calculated is large, the molecular orbital energies calculated will resemble Diagram #1. Calculations are an important part of the discussion.

Both diagrams predict an increase in ionization potential on going from atomic N to molecular N₂ (actual ionization potential for N is 1402 kJ/mol; for N₂ is 1503 kJ/mol), but the difference does not allow a definitive choice between the models.

So it comes down to estimating the hybridization and doing a calculation. I suspect that if you try to determine something about the first ionized state of N₂, it would resemble Diagram #1 more than #2, because the nitrogen atoms would be drawn closer together.

Another differentiation would be to see how nitrogen molecules bind to other atoms. Ref 2 indicates that the bonding is linear, in agreement with both diagrams. Nitrogen sometimes (rarely) bonds with two atoms (a side bond). The question allows for a lot of pictorial discussion, but not much in a numerical sense.

It could be looked at as a comparison between a stabilizing, but formally sigma antibond against a pi bond and an interesting way to bring up hybridization and overlaps.

Ref 1. In molecular orbital theory, why does s-p mixing in the dinitrogen molecule not effect the 1σᵤ orbital?

Ref 2. https://en.wikipedia.org/wiki/Transition_metal_dinitrogen_complex